{"id":2096,"date":"2022-02-22T13:03:58","date_gmt":"2022-02-22T04:03:58","guid":{"rendered":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/?page_id=2096"},"modified":"2024-04-20T15:43:29","modified_gmt":"2024-04-20T06:43:29","slug":"%e4%b8%89%e8%a7%92%e9%96%a2%e6%95%b0%e3%81%ae%e5%be%ae%e5%88%86","status":"publish","type":"page","link":"https:\/\/home.hirosaki-u.ac.jp\/relativity\/%e7%90%86%e5%b7%a5%e7%b3%bb%e3%81%ae%e6%95%b0%e5%ad%a6b\/%e4%b8%89%e8%a7%92%e9%96%a2%e6%95%b0%e3%81%ae%e5%be%ae%e5%88%86\/","title":{"rendered":"\u4e09\u89d2\u95a2\u6570\u306e\u5fae\u5206"},"content":{"rendered":"

\u5f27\u5ea6\u6cd5\uff08\u30e9\u30b8\u30a2\u30f3\u5358\u4f4d\uff09\u3067 \\(x\\) \u3092\u8868\u3059\u3068\uff0c
\n$$(\\sin x)’ = \\cos x, \\quad (\\cos x)’ = -\\sin x, \\quad (\\tan x)’ = \\frac{1}{\\cos^2 x}$$<\/p>\n

\u3068\u306a\u308b\u3053\u3068\u3092\u793a\u3059\u3002<\/p>\n

\n

\u4e09\u89d2\u95a2\u6570\u306e\u5b9a\u7fa9<\/h3>\n

\u56f3\u306e\u3088\u3046\u306a\u534a\u5f84 \\(r\\) \u306e\u5186\u4e0a\u306e\u70b9 \\(P({\\color{blue}{x}}, {\\color{red}{y}})\\) \u3092\u8003\u3048\uff0c\\(x\\) \u8ef8\u304b\u3089\u306e\uff08\u53cd\u6642\u8a08\u56de\u308a\u3092\u6b63\u306e\u5411\u304d\u3068\u3057\uff0c\u30e9\u30b8\u30a2\u30f3\u3067\u8868\u3057\u305f\uff09\u89d2\u5ea6\u3092 \\({\\color{green}{\\theta}}\\) \u3068\u3059\u308b\u3068\uff0c\u4e09\u89d2\u95a2\u6570\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u5b9a\u7fa9\u3055\u308c\u308b\u306e\u3067\u3042\u3063\u305f\u3002<\/p>\n

$$
\n\\cos{\\color{green}{\\theta}} = \\frac{\\color{blue}{x}}{r}, \\quad
\n\\sin{\\color{green}{\\theta}} = \\frac{\\color{red}{y}}{r}, \\quad
\n\\tan{\\color{green}{\\theta}} = \\frac{\\color{red}{y}}{\\color{blue}{x}} = \\frac{\\sin{\\color{green}{\\theta}}}{\\cos{\\color{green}{\\theta}}}
\n$$<\/p>\n

\"\"<\/a><\/p>\n

\u4e09\u89d2\u95a2\u6570\u306e\u57fa\u672c\u516c\u5f0f<\/h3>\n

$$\\cos^2\\theta + \\sin^2\\theta = 1$$<\/p>\n

\u5ff5\u306e\u305f\u3081\uff0c\\(\\cos^2\\theta = \\left( \\cos\\theta\\right)^2, \\ \\sin^2\\theta = \\left( \\sin\\theta\\right)^2\\) \u306e\u3053\u3068\u3067\u3059\u3002<\/p>\n

\u3053\u308c\u306f\u76f4\u89d2\u4e09\u89d2\u5f62\u306b\u5bfe\u3059\u308b\u30d4\u30bf\u30b4\u30e9\u30b9\u306e\u5b9a\u7406\uff08\u4e09\u5e73\u65b9\u306e\u5b9a\u7406\uff09<\/p>\n

$$ x^2 + y^2 = r^2$$<\/p>\n

\u306e\u4e21\u8fba\u3092 \\(r^2\\) \u3067\u5272\u3063\u3066\u4e09\u89d2\u95a2\u6570\u3067\u66f8\u304d\u306a\u304a\u3057\u305f\u3082\u306e\u306b\u4ed6\u306a\u3089\u306a\u3044\u3002<\/p>\n

\u307e\u305f\uff0c\u4e09\u89d2\u95a2\u6570\u306e\u52a0\u6cd5\u5b9a\u7406\u306b\u983c\u308b\u307e\u3067\u3082\u306a\u304f\uff0c\u4ee5\u4e0b\u306e\u3053\u3068\u306f\u4e0a\u56f3\u306e\u76f4\u89d2\u4e09\u89d2\u5f62\u3092\u300c\u6a2a\u5012\u3057\u300d\u306b\u3057\u3066\u773a\u3081\u3066\u307f\u308c\u3070\u308f\u304b\u308b\u3060\u308d\u3046\u3002<\/p>\n

$$\\cos\\left( {\\color{purple}{\\frac{\\pi}{2}\u00a0 -\\theta}}\\right) = \\frac{\\color{red}{y}}{r} = \\sin {\\color{green}{\\theta}}, \\quad \\sin\\left( {\\color{purple}{\\frac{\\pi}{2}\u00a0 -\\theta}}\\right)\u00a0 = \\frac{\\color{blue}{x}}{r} = \\cos {\\color{green}{\\theta}}$$<\/p>\n

\"\"<\/a><\/p>\n

\u5fae\u5206\u3092\u793a\u3059\u305f\u3081\u306b\u5fc5\u8981\u306a\u516c\u5f0f<\/h3>\n

\u4e09\u89d2\u95a2\u6570\u306e\u5fae\u5206\u3092\u793a\u3059\u305f\u3081\u306b\u5fc5\u8981\u306a\u306e\u306f\uff0c\u6b21\u306e4\u3064\u3002\u3053\u308c\u3089\u3092\u7c21\u5358\u306b\u8aac\u660e\u3002<\/p>\n

\u307e\u305a\uff0c\u4e09\u89d2\u95a2\u6570\u306e\u6975\u9650\u516c\u5f0f<\/p>\n

1.\u00a0 \\(\\quad\\displaystyle \\lim_{x \\rightarrow 0} \\frac{\\sin x}{x} = 1 \\)<\/p>\n

2.\u00a0 \\(\\quad\\displaystyle \\lim_{x \\rightarrow 0} \\frac{1-\\cos x}{x} = 0 \\)<\/p>\n

\u8a3c\u660e\u306f\uff0c1. \u3092\u4f7f\u3063\u3066\uff0c
$$ \\lim_{x \\rightarrow 0}\\frac{\\sin^2 x}{x} = \\lim_{x\\rightarrow 0} \\sin x\\cdot \\lim_{x\\rightarrow 0} \\frac{\\sin x}{x} = 0\\cdot 1 = 0$$<\/p>\n

\u4e00\u65b9\uff0c
\\begin{eqnarray}
\\lim_{x \\rightarrow 0}\\frac{\\sin^2 x}{x} &=&
\\lim_{x \\rightarrow 0}\\frac{(1+\\cos x)(1-\\cos x)}{x} \\\\
&=& \\lim_{x \\rightarrow 0}(1+\\cos x)\\cdot \\lim_{x \\rightarrow 0}\\frac{1-\\cos x}{x}\\\\
&=& 2 \\lim_{x \\rightarrow 0}\\frac{1-\\cos x}{x}
\\end{eqnarray}
\u3057\u305f\u304c\u3063\u3066
$$\\lim_{x \\rightarrow 0}\\frac{1-\\cos x}{x} = 0$$<\/p>\n

\u305d\u3057\u3066\uff0c\u4e09\u89d2\u95a2\u6570\u306e\u52a0\u6cd5\u5b9a\u7406\uff1a<\/p>\n

3.\u00a0 \\(\\displaystyle \\quad \\sin (x + h) = \\sin x \\cos h + \\cos x \\sin h \\)<\/p>\n

4.\u00a0\u00a0\\(\\displaystyle \\quad \\cos(x+h) = \\cos x \\cos h -\\sin x \\sin h \\)<\/p>\n

\u5b9a\u7fa9\u304b\u3089\u5c0e\u304f\u4e09\u89d2\u95a2\u6570\u306e\u5fae\u5206<\/h3>\n

\u5c0e\u95a2\u6570\u306e\u5b9a\u7fa9\u304b\u3089 \\( (\\sin x)’\\) \u306f\uff0c<\/p>\n

\\begin{eqnarray}
(\\sin x)’ &=& \\lim_{h \\rightarrow 0} \\frac{\\sin(x + h) -\\sin x}{h} \\\\
&=& \\lim_{h \\rightarrow 0} \\frac{(\\sin x \\cos h + \\cos x \\sin h) -\\sin x}{h}\\\\
&=& \\sin x\\cdot\\lim_{h \\rightarrow 0}\\frac{(\\cos h -1)}{h} + \\cos x\\cdot\\lim_{h \\rightarrow 0}\\frac{\\sin h}{h}\\\\
&=& \\sin x \\cdot 0 + \\cos x \\cdot 1\\\\
&=& \\cos x
\\end{eqnarray}<\/p>\n

\\( (\\cos x)’\\) \u3082\u540c\u69d8\u306b
\\begin{eqnarray}
(\\cos x)’ &=& \\lim_{h \\rightarrow 0} \\frac{\\cos(x + h) -\\cos x}{h} \\\\
&=& \\lim_{h \\rightarrow 0} \\frac{(\\cos x \\cos h -\\sin x \\sin h)-\\cos x}{h}\\\\
&=& \\cos x\\cdot\\lim_{h \\rightarrow 0}\\frac{(\\cos h -1)}{h} -\\sin x\\cdot\\lim_{h \\rightarrow 0}\\frac{\\sin h}{h}\\\\
&=& \\cos x \\cdot 0 -\\sin x \\cdot 1\\\\
&=& -\\sin x
\\end{eqnarray}<\/p>\n

\\( (\\sin x)’ \\) \u306e\u5fae\u5206\u3060\u3051\u899a\u3048\u3066\u304a\u3051\u3070\uff0c\u4ed6\u306f\u3059\u3079\u3066\u5c0e\u3051\u308b\u3093\u3060\u3068\u3044\u3046\u7acb\u5834\u3092\u597d\u3080\u4eba\u306a\u3089\u3070\uff0c\\( \\displaystyle\\cos x = \\sin\\left(\\frac{\\pi}{2} -x\\right) \\) \u3092\u4f7f\u3044\uff0c\\(u \\equiv \\displaystyle \\frac{\\pi}{2} -x\\) \u3068\u304a\u3044\u3066\u5408\u6210\u95a2\u6570\u306e\u5fae\u5206\u3092\u4f7f\u3063\u3066\u8a3c\u660e\u3057\u307e\u3057\u3087\u3046\u3002
\\begin{eqnarray}
(\\cos x)’ &=& \\frac{d}{dx} \\sin\\left(\\frac{\\pi}{2} -x\\right) \\\\
&=& \\frac{d\\sin u}{du} \\frac{du}{dx} \\\\
&=& \\cos \\left(\\frac{\\pi}{2} -x\\right) \\cdot (-1) \\\\
&=& -\\sin x
\\end{eqnarray}<\/p>\n

\\( (\\tan x)’ \\) \u306b\u3064\u3044\u3066\u306f\uff0c\u5fae\u5206\u6cd5\u306e\u516c\u5f0f 5. \u3088\u308a
$$ (\\tan x)’ = \\left\\{ \\frac{\\sin x}{\\cos x} \\right\\}’\u00a0 = \\frac{(\\sin x)’ \\cos x -\\sin x (\\cos x)’}{\\cos^2 x} = \\cdots$$
\u3068\u3057\u3066\uff0c\\( \\cos^2 x + \\sin^2 x = 1\\) \u3092\u4f7f\u3046\u306e\u3067\u3042\u3063\u305f\u3002<\/p>\n

\u4e09\u89d2\u95a2\u6570\u306e\u30b0\u30e9\u30d5<\/h3>\n

3\u3064\u307e\u3068\u3081\u3066\u30b0\u30e9\u30d5\u306b\u3059\u308b\u3068…<\/p>\n

\"\"<\/p>\n

<\/div>\n","protected":false},"excerpt":{"rendered":"

\u5f27\u5ea6\u6cd5\uff08\u30e9\u30b8\u30a2\u30f3\u5358\u4f4d\uff09\u3067 \\(x\\) \u3092\u8868\u3059\u3068\uff0c $$(\\sin x)’ = \\cos x, \\quad (\\cos x)’ = -\\sin x, \\quad (\\tan x)’ = \\frac{1}{\\cos^2 x}$$<\/p>\n

\u3068\u306a\u308b\u3053\u3068\u3092\u793a\u3059\u3002<\/p>

\u7d9a\u304d\u3092\u8aad\u3080<\/a><\/p>\n","protected":false},"author":33,"featured_media":0,"parent":2068,"menu_order":6,"comment_status":"closed","ping_status":"closed","template":"","meta":{"inline_featured_image":false,"footnotes":""},"class_list":["post-2096","page","type-page","status-publish","hentry","nodate","item-wrap"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t